Is the ring of invariants Noetherian?

Question

Let $R$ be a complete regular local ring whose residue field is perfect. Suppose that a finite group $G$ acts on $R$ by ring automorphisms in such a way that the induced action on the residue field is trivial. Is the ring of invariants $R^G$ necessarily Noetherian?

Answer 1

Yes, and regularity isn't needed (assuming noetherian). By the Eakin-Nagata Theorem (3.7, Matsumura CRT), it is enough that $R$ is $R^G$-finite. For the Cohen ring $W$ of the perfect residue field, the unique local map $W\rightarrow R$ lifting the identity on residue fields is $G$-invariant. Pick a surjection $W[\![x_1, \dots, x_n]\!]\rightarrow R$. Let $t_{i1},\dots$ in $R^G$ be the indexed sequence of $\#G$ elementary symmetric "functions" in the $G$-orbit of $x_i$. Then the unique local $W$-algebra map $W[\![T_{ij}]\!] \rightarrow R$ carrying $T_{ij}$ to $t_{ij}$ is module-finite by completeness of $R$ since each $x_i$ is nilpotent in $R/(T_{ij})R$, yet this factors through $R^G$, so $R$ is also $R^G$-finite. QED